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Unformatted text preview: Solutions to the second midterm exam (November 14)
Problem 1. Matrix A deﬁnes a linear transformation R4 −→ R3 , x −→ Ax. Hence the
kernel of A lies in R4 and the image of A lies in R3 . Bringing A to the reduces row echelon
form, we obtain 113
2 1 5
317 4
11
5 −→ 0 −1
6
0 −2 3
−1
−2 4
11
−3 −→ 0 1
−6
00 34
1
1 3 −→ 0
00
0 021
1 1 3.
000 Hence the kernel of A consists of the vectors −1
−2
−2x3 − x4 −3 −1 −x3 − 3x4 , + x4 = x3 0
1
x3
1
0
x4 −1 −2 −1 −3 where x3 and x4 can be any numbers. We conclude that is a basis of
,
0
1 1
0
the kernel A. Selecting the columns of A corresponding to the leading 1’s, we conclude of 1
1 2 , 1 is a basis of the image of A.
that 3
1 −1 −2 −1 −3 ,
Answer: Vectors form a basis of the kernel of A. Geometrically, the 0
1 1
0 1
1
4 2 , 1 form a basis of the image of A.
kernel of A is a plane in R . Vectors 3
1
Geometrically, the image of A is a plane in R3 .
Problem 2. The cosine of the angle between a + b and a − b is
a+b · a−b . a−b a+b
Now, a+b · a−b =a·a+b·a−a·b−b·b = a
1 2 2 −b = 4 − 9 = −5, a+b =
= a+b · a+b =
a 2 a·a+b·a+a·b+b·b 2 + 2a · b + b = √ 9+4+4 = √ 17 and
a−b =
= a−b · a−b =
a 2 a·a−b·a−a·b+b·b 2 − 2a · b + b = √ 9 − 4 + 4 = 3. √
Hence the cosine of the angle is −5/3 17. 5
Answer: The cosine of the angle between a + b and a − b is − √ .
3 17 Problem 3. Matrix A deﬁnes a linear transformation R3 −→ R5 , x −→ Ax, while matrix
B deﬁnes a linear transformation R4 −→ R3 , x −→ Bx.
a) Since dim ker A + dim im A = 3 and dim im A = rank A = 2, we conclude that
dim ker A = 1.
Answer: the dimension of the kernel of A is 1.
b) Since dim ker B + dim im B = 4 and dim im B = rank B = 2, we conclude that
dim ker B = 2.
Answer: the dimension of the kernel of B is 2.
c) We have (AB )x = A (Bx) for all x in R4 . The dimension of the image of B is 2,
so, geometrically the image of B is a plane in R3 . Now, the kernel of A is a line. If the
kernel of A lies in the image of B then the dimension of the image of AB will be 1, so the
dimension of the kernel of AB will be 4 − 1 = 3. If the kernel of A intersects the image of
B in the origin only, the dimension of the image of AB will be 2, so the dimension of the
kernel of AB will be 4 − 2 = 2. ker (A) im( B ) im(AB) ker (A)
im( B )
2 im(AB) In fact, both values can occur, for example, when
1
0 A = 0 0
0 0
1
0
0
0 and the dimension 10
0 1 A = 0 0 00
00 0
0 0 0
0 and 10
B = 0 1
00 of the kernel of AB is 0 0
0 0 and B = 1 0
0
0 00
0 0
00 then 1
0 AB = 0 0
0 0
1
0
0
0 0
0
0
0
0 0
0 0 0
0 then 0
1 AB = 0 0
0 0
0
0
0
0 0
0
0
0
0 0
0 0 0
0 2. If 0
0
1 00
0 0
00 and the dimension of the kernel of AB is 3.
Answer: the possible dimensions of the kernel of AB are 2 and 3.
Problem 4. Writing the system in the form Ax = b, where 1 −1
0 2 −1 , b = 1 and x = x ,
A=
y
31
3
We solve the system AT Ax = AT b, where
AT A = AT b = 1
−1 2
−1 1
2
−1 −1 3
1
3
1 1 −1 2 −1 = 14 0
03
31 0 1 = 11 ,
2
3 and that is, 14x = 11 and 3y = 2, from which x = 11/14 and y = 2/3.
Answer: the least squares solution is x = 11/14 and y = 2/3.
Problem 5.
a) We look for the projection w in the form w = xa + y b for some x and y such that
(v − w) · a = 0 and (v − w) · b = 0. That is, 1
1
0
1 0 − x 1 − y 1 · 1 = 0
1
0
1
0 and 3 1
1
0
0 0 − x 1 − y 1 · 1 = 0.
1
0
1
1 Computing the dot products, we obtain
2x + y = 1 and x + 2y = 1, from which x = y = 1/3 and 1
0
1/3
1
1
w = 1 + 1 = 2/3 .
3
3
0
1
1/3 1/3
Answer: the orthogonal projection is 2/3 .
1/3
b) Let u be the orthogonal projection of v onto the orthogonal complement to the
plane spanned by a and b. u v
w Then u + w = v , so 1
1/3
2/3
u = v − w = 0 − 2/3 = −2/3 .
1
1/3
2/3 2/3
Answer: the orthogonal projection is −2/3 .
2/3 4 ...
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This note was uploaded on 12/26/2011 for the course MATH 214 taught by Professor Conger during the Fall '08 term at University of Michigan.
 Fall '08
 Conger
 Linear Algebra, Algebra

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